Three forms of mathematical induction

This article presumes familiarity with mathematical induction. This is not the appropriate place to learn what mathematical induction is or how to write proofs by mathematical induction. The article titled mathematical induction, not this one, is the appropriate source of that kind of information.

The three forms

Proofs that a subset of { 1, 2, 3, ... } is in fact the whole set { 1, 2, 3, ... } by mathematical induction usually have one of the following three forms. They are collected here in order to show the contrast between them, and its coexistence with the commonality between them.

First form

• The basis for induction is trivial;
• the substantial part of the proof goes from case n to case n + 1.

Second form

• The first case is vacuously true;
• the step that goes from the nth case to the (n + 1)th case is trivial if n ≥ 2 and impossible if n = 1;
• the substantial part of the proof is the second case;
• the second case is relied on in the trivial induction step.

Third form

• The induction step shows that if P(k) is true for all k < n then P(n) is true (proof by complete induction);
• no basis for induction is needed because the first, or basic, case is a vacuously true special case of what is proved in the induction step.

This form works not only when the values of k and n are natural numbers, but also for ordinal numbers; see transfinite induction.

Examples

This is not the appropriate place to go into details of proofs, but rather only broad outlines of proofs are shown. For examples of proofs by mathematical induction with full details, see mathematical induction.

First form

Most proofs by mathematical induction that are adduced as examples when mathematical induction is first taught in secondary school are of the first form.

Second form

Product rule

The usual product rule taught in calculus says

${\displaystyle (fg)'=f'g+g'f.\,}$ 

For a product of n functions, one has

${\displaystyle (f_{1}f_{2}f_{3}\cdots f_{n})'\,}$ 

${\displaystyle =(f_{1}'f_{2}f_{3}\cdots f_{n})+(f_{1}f_{2}'f_{3}\cdots f_{n})+(f_{1}f_{2}f_{3}'\cdots f_{n})+\cdots +(f_{1}f_{2}\cdots f_{n-1}f_{n}').\,}$ 

In each of the n terms, just one of the factors is a derivative; the others are not.

Now observe four facts:

• If n = 1, then the proposition just says

${\displaystyle f'=f'.\,}$ 

It can be said that just one factor is a derivative, and it is vacuously true that the others are not, because there are no others!

• To go from case n to case n + 1 is trivial if n ≥ 2, but impossible if n = 1.

• To go from case n to case n + 1, one must use the case n = 2 (i.e. one must use the ordinary product rule taught in differential calculus).

• The substantial part of the proof is the case n = 2, and that is precisely the product rule that is taught differential calculus.

Triangle inequality

The usual triangle inequality in metric spaces says

${\displaystyle d(x_{1},x_{3})\leq d(x_{1},x_{2})+d(x_{2},x_{3}).\,}$ 

For a sequence of n points, one has

${\displaystyle d(x_{1},x_{n})\leq d(x_{1},x_{2})+\cdots +d(x_{n-1},x_{n}).\,}$ 

Now observe four facts

• The first case is

${\displaystyle d(x_{1},x_{2})\leq d(x_{1},x_{2}),\,}$ 

and this is vacuously true, in that it does not rely on any information about what kind of function d is (i.e., that d is a metric).

• The nth case is the case with n terms on the right side, and that is

${\displaystyle d(x_{1},x_{n+1})\leq d(x_{1},x_{2})+\cdots +d(x_{n},x_{n+1}).\,}$ 

To go from there to the (n + 1)th case is trivial if n ≥ 2, but impossible if n = 1.

• To go from case n to case n + 1, one must use the case n = 2 (i.e. one must use the ordinary triangle inequality).

• The substantial part of the proof is the second case. The second case is just the statement of the ordinary triangle inequality, and is different in different metric spaces. It is part of the proof that a particular function d is in fact a metric. In some cases is difficult or otherwise onerous. Usually the other aspects of proving that d is a metric are trivial (i.e. that d(x, x) = 0 for all x, and that d is symmetric).

Polya's proof that there is no "horse of a different color"

In the middle of the 20th century, a commonplace colloquial locution to express the idea that something is unexpectedly different from the usual was "That's a horse of a different color!". George Polya posed the following exercise: find the error in the following argument, which purports to prove by mathematical induction that all horses are of the same color:

If there's only one horse, there's only one color.

Suppose within any set of n horses, there is only one color. Now look at any set of n + 1 horses. Number them: 1, 2, 3, ..., n, n + 1. Consider the sets {1, 2, 3, ..., n} and {2, 3, 4, ..., n + 1}. Each is a set of only n horses, therefore with each there is only one color. But the two sets overlap, so there must be only one color among all n + 1 horses.

Now observe four facts

• The first case— with only one horse—is and this is vacuously true; there are no counterexamples—two horses of different colors—because there are no examples at all of two horses.

• To go from the case of n horses to the case of n + 1 horses is trivial if n ≥ 2 (it is done above), but impossible if n = 1.

(Spoiler: the impossibility when n = 1 is precisely the error that Polya challenged the student to find).

• To go from case n to case n + 1, one must use the fact that any two horses within either of the two sets of n are of the same color, i.e., one must use the case n = 2.

• The substantial part of the proof is the second case, i.e. the hard part is to prove that any two horses are of the same color. Once you've got that, it's easy to see that any set of more than two are of the same color.